hwsolB02

# hwsolB02 - Therefore v^(13 = v^-1 = v(complex conjugate of...

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Sheet1 Page 1 HW 2 SOLUTION _____________ (i) The polar coordinates are |u| = 1, angle(u) = pi/7 |v| = 1, angle(v) = 6*pi/7 |w| = 1, angle(u) = 2*pi/5 (ii) |z-u| = |z-v| is tha perpendicular bisector of the segment joining u and v. This straight line is the y-axis (x = 0) (iii) |z-u-u^2*v|=|w| is a circle of radius |w| = 1 with center at the point u+u^2*v = 0+0j (Origin). All points u, v, and w lie on the circle. (iv) Here we have to work with Cartesian forms. v = cos(2*pi/7) + j*sin(2*pi/7) = 0.6235 + j*0.7818 w = cos(2*pi/5) + j*sin(2*pi/5)) = 0.3090 + j*0.9511 Thus v+c*w is purely real when 0.6235 + c*0.3090 = 0, i.e., c = -0.4956 purely imaginary when 0.7818 + c*0.9511 = 0, i.e., c = -1.2166 (v) v^13 = exp(j*13*6*pi/7) = exp(j*(12*pi - 6*pi/7)) = exp(-j*6*pi/7)

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Unformatted text preview: Therefore v^(13) = v^(-1) = v' (complex conjugate of v). We know that exp(j*theta)+exp(-j*theta) = 2*cos(theta) exp(j*theta)-exp(-j*theta) = 2*j*sin(theta) and thus v + v^13 = 2*cos(6*pi/7) v - v^13 = 2*j*sin(6*pi/7) (vi) The polynomial is z^5-1 . Based on our discussion, z^5 = 1 has z = 1 as its root, along with four more roots: exp(j*2*k*pi/5) for k=1,2,. ..,4. This is the same as the set w^5=1, w, w^2, w^3, and w^4 Sheet1 Page 2 (vii) The geometric sum formula gives 1 + z + z^2 + . .. + z^(n-1) = (1 - z^n)/(1 - z) for all z except z=1. Setting z=v, we have 1 + w + w^2 + w^3 + w^4 = (1 - w^5)/(1 - w) But w^5 = 1, and thus 1 + w + w^2 + w^3 + w^4 = 0...
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## This note was uploaded on 03/21/2010 for the course ENEE 241 taught by Professor Staff during the Spring '08 term at Maryland.

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hwsolB02 - Therefore v^(13 = v^-1 = v(complex conjugate of...

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